Вычислите: 1. [latex] \frac{log _{4} 36}{log _{4}6 } [/latex] 2. [latex] \frac{log _{0,3}32 }{log _{0,3} 64} [/latex] 3. [latex] \frac{2log _{0,5}2+log _{0,5} \sqrt{10} }{ log_{0,5} 10- log_{0,5} \sqrt{10} + log_{0,5} 4} [/lat...
Вычислите:
1. [latex] \frac{log _{4} 36}{log _{4}6 } [/latex]
2. [latex] \frac{log _{0,3}32 }{log _{0,3} 64} [/latex]
3. [latex] \frac{2log _{0,5}2+log _{0,5} \sqrt{10} }{ log_{0,5} 10- log_{0,5} \sqrt{10} + log_{0,5} 4} [/latex]
4. [latex] \frac{ log_{0,3}16 }{ log_{0,3} 15- log_{0,3} 30} [/latex]
Ответ(ы) на вопрос:
[latex] \frac{log _{4}36}{log _{4}6} = \frac{log _{4}6^2}{log _{4}6} = \frac{2log _{4}6}{log _{4}6} = 2;[/latex]
[latex] \frac{log _{0,3}32}{log _{0,3}64} = \frac{log _{0,3}2^5}{log _{0,3}2^6} = \frac{5log _{0,3}2}{6log _{0,3}2} = \frac{5}{6}; [/latex]
[latex] \frac{2log _{0,5}2+log _{0,5}\sqrt{10}}{log_{0,5}10-log_{0,5}\sqrt{10}+log_{0,5}4} = \frac{log _{0,5}(2^2\cdot\sqrt{10})}{log_{0,5} \frac{10\cdot4}{\sqrt{10}} } = \frac{log _{0,5}(4\cdot\sqrt{10})}{log _{0,5}(4\cdot\sqrt{10})} = 1;[/latex]
[latex] \frac{log_{0,3}16}{log_{0,3}15-log_{0,3}30} = \frac{log_{0,3}2^4}{log_{0,3}\frac{15}{30}} = \frac{4log_{0,3}2}{log_{0,3}2^{-1}} = \frac{4log_{0,3}2}{-log_{0,3}2}} = -4.[/latex]
Не нашли ответ?
Похожие вопросы